L 3, -Solutions to the Navier-Stokes Equations and Backward Uniqueness

L 3, -Solutions to the Navier-Stokes Equations and Backward Uniqueness L. Escauriaza, G. Seregin, V. Šverák Dedicated to Olga Alexandrovna Ladyzhenskaya Abstract We show that L 3, -solutions to the Cauchy problem for the threedimensional Navier-Stokes equations are smooth. 1991 Mathematical subject classification (Amer. Math. Soc.): 35K, 76D. Key Words: the Navier-Stokes equations, the Cauchy problem, weak Leray- Hopf solutions, suitable weak solutions, backward uniqueness. Table of contents 1.Introduction..Suitable weak solutions. 3.Proof of the main results. 4.Unique continuation through spatial boundaries. 5.Backward uniqueness. 6.Carleman-type inequalities. 7.Appendix. 1 Introduction In this expository paper, we review some recent results in the regularity theory for the Navier-Stokes equations. We consider the classical Cauchy problem for these equations: t v(x, t) + div v(x, t) v(x, t) v(x, t) = p(x, t), div v(x, t) = 0 (1.1) 1

for x R 3 and t 0, together with the initial condition v(x, 0) = a(x), x R 3. (1.) We assume for the moment that a is a smooth divergence-free vector field in R 3 which decays sufficiently fast as x. (We will return later to the important case when the initial velocity field a belongs to more general classes of functions.) In the classical paper [19], Leray proved the following results: (i) There exists a T > 0 such that the Cauchy problem (1.1), (1.) has a unique smooth solution with reasonable properties at. (ii) Problem (1.1), (1.) has at least one global weak solution satisfying a natural energy inequality. Moreover, the weak solutions coincide with the smooth solution in R 3 ]0, T [. (iii) If ]0, T [ is the maximal interval of the existence of the smooth solution, then, for each p > 3, there exists ε p > 0 such that ( R 3 u(x, t) p dx ) 1 p ε p (T t) 1 (1 3 p ) as t T. (iv) For a given weak solution, there exists a closed set S ]0, + [ of measure zero such that the solution is smooth in R 3 (]0, [\S). (In fact, Leray s proof gives us S with H 1 (S) = 0, although it is not mentioned explicitly.) The modern definition of the weak solutions (often called Leray-Hopf weak solutions due to important contributions of E. Hopf in the case of bounded domains) is as follows. We denote by C 0 the space of all infinitely differentiable solenoidal vector fields with compact support in R 3 ; J and J 1 are the closures of the set C 0 in the spaces L and W 1, respectively. (We use the standard notation for the Lebesgue and Sobolev spaces.) In what follows we will use the notation Q T = R 3 ]0, T [. A Leray-Hopf weak solution of the Cauchy problem (1.1) and (1.) in Q T is a vector field v : Q T IR 3 such that v L (0, T ; J) L (0, T ; J 1 ); (1.3) the function t v(x, t) w(x) dx can be continuously extended R3 to [0, T ] for any w L ; (1.4)

( v t w v v : w + v : w) dxdt = 0, w C 0 (Q T ); (1.5) Q T 1 v(x, t 0 ) dx+ v dxdt 1 a(x) dx, t 0 [0, T ]; (1.6) R3 R3 ]0,t 0 [ R3 v(, t) a( ) 0 as t 0. (1.7) The definition is meaningful also for T = + if we replace the closed interval [0, T ] by [0, [ throughout the definition. Leray s result (ii) above can now be stated as follows. (See [19], [1], [14], and [16].) Theorem 1.1 Assume that a J. (1.8) Then there exists at least one Leray-Hopf weak solution to the Cauchy problem (1.1) and (1.) in R 3 ]0, [. At the time of this writing, both uniqueness and regularity of Leray-Hopf weak solutions remain open problems. Important extensions of Leray s results were later obtained by many workers. In particular, the works of Prodi [30], Serrin [4], and Ladyzhenskaya [15] lead to the following generalizations of (ii). Theorem 1. Suppose that condition (1.8) holds. Let v and v 1 be two weak Leray-Hopf solutions to the Cauchy problem (1.1) and (1.). Assume that, for some T > 0 the velocity field v satisfies the so-called Ladyzhenskaya- Prodi-Serrin condition, i.e., v L s,l (Q T ) (1.9) with 3 s + = 1, s ]3, + ]. (1.10) l Then, v = v 1 in Q T and, moreover, v is a smooth function in R 3 (0, T ]. The uniqueness was proved by Prodi in [30] and Serrin in [4] and the smoothness was established by Ladyzhenskaya in [15]. Further extension of Theorem 1. can be found in paper of Giga [10]. A local version of this theorem was 3

proved by Serrin [41] for 3 + < 1 and Struwe [45] for 3 + = 1. We recall s l s l that the norm in the mixed Lebesgue space L s,l (Q T ) is given as follows: ( T ) 1 f(, t) l l s dt, l [1, + [ 0 f s,l,qt = ess sup f(, t) s, l = +. t ]0,T [ If s = l, we abbreviate f s,qt = f s,s,qt. We note that, by standard imbeddings, functions of the Leray-Hopf class satisfy v L s,l (Q T ) (1.11) with 3 s + l = 3, s [, 6]. (1.1) Hence there is a substantial gap between what we have according to the existence theorem and what we need for uniqueness. An important step towards understanding regularity properties of the weak Leray-Hopf solutions was a localization in x of Leray s results (iv). This program was started by Scheffer [31] [34] and developed further by Caffarelli-Kohn-Nirenberg []. Recently, Lin [0] outlined significant simplifications in the proof of these results (see also [17] for more detail proofs). In this paper, we address the problem of regularity for the weak Leray- Hopf solutions v satisfying the additional condition v L 3, (Q T ). (1.13) We prove that Leray s result (iii) has the following analogue for p = 3. If ]0, T [ is the maximal interval of the existence of the smooth solution to problem (1.1), (1.) and T < +, then lim sup v(x, t) 3 dx = +. t T R 3 In other words, the spatial L 3 -norm of v must blow-up if the solution develops a singularity. We can also view this result as an extension of Theorem 1. to the case s = 3, l = +. More precisely, we have 4

Theorem 1.3 Assume that v is a weak Leray-Hopf solution to the Cauchy problem (1.1) and (1.) in Q T and satisfies the additional condition (1.13). Then, v L 5 (Q T ) (1.14) and hence it is smooth and unique in Q T. The uniqueness of v under condition (1.13) has already been known, see [, 3, 6, 43]. Some partial results on smoothness of such solutions can be found in [1, 9]. In fact, we prove the following local result. Theorem 1.4 Consider two functions v and p defined in the space-time cylinder Q = B ] 1, 0[, where B(r) R 3 stands for the ball of radius r with the center at the origin and B = B(1). Assume that v and p satisfy the Navier-Stokes equations in Q in the sense of distributions and have the following differentiability properties: Let, in addition, v L, (Q) L ( 1, 0; W 1 (B)), p L 3 (Q). (1.15) v 3,,Q < +. (1.16) Then the function v is Hölder continuous in the closure of the set Q(1/) = B(1/) ] (1/), 0[. The main interest of the above results comes from the fact that they seem out to be of reach of standard methods. By those methods, we mean various conditions on (local) smallness of various norms of v which are invariant with respect to the natural scaling u(x, t) λu(λx, λ t), p(x, t) λ p(λx, λ t) of the equations. We note that finiteness of a norm f s,l with s, l < implies local smallness of f in this norm. This is not the case for L 3, -norm (which is still invariant under the scaling). This possible concentration effect was the main obstacle to proving regularity. To rule out concentration, we use a new method based on the reduction of the regularity problem to a backward uniqueness problem, which is than solved by finding suitable Carleman-type 5

inequalities. The backward uniqueness results are new and seem to be of independent interest, see Section 5 and Section 6. Our methods can be probably easily adopted to other parabolic problems with critical non-linearities. In fact, one could speculate that the general idea of the approach might be applicable to an even larger class of interesting equations with critical non-linearities, such as non-linear Schrödinger equations or non-linear wave equations. However, the local regularity issues arising in these cases would be slightly harder than in the parabolic case. The plan of the paper is as follows. In Section, we discuss known results about regularity of so-called suitable weak solutions. In the third section, we reduce the regularity problem to the backward uniqueness for the heat operator with variable lower order terms. This proves Theorem 1.4 and therefore Theorem 1.3. In Section 4, we discuss known facts from the theory of the unique continuation of solutions to parabolic equations through spatial boundaries. In the next section, we prove the backward uniqueness result used in Section 3. The sixth section is devoted to the derivation of two Carleman-type inequalities, which play the crucial role in our proof of the backward uniqueness theorem. Finally, just for completeness, we present the known theorem on the short time solvability of the Cauchy problem with the initial data from L 3 J in the class C([0, T ]; L 3 ) L 5 (Q T ) in the Appendix. Suitable Weak Solutions In this section, we are going to discuss smoothness of the so-called suitable weak solutions to the Navier-Stokes equations. The definition of suitable weak solutions was introduced in [], see also [31]-[34], [0], and [17]. Our version is due to [17]. Definition.1 Let ω be a open set in R 3. We say that a pair u and q is a suitable weak solution to the Navier-Stokes equations on the set ω ] T 1, T [ if it satisfies the conditions: u L, (ω ] T 1, T [) L ( T 1, T ; W 1 (ω)); (.1) q L 3 (ω ] T 1, T [); (.) u and q satisfy the Navier-Stokes equations in the sense in distributions; 6 (.3)

u and q satisfy the local energy inequality ϕ(x, t) u(x, t) dx + ω ω ] T 1,t[ ω ] T 1,t[ ϕ u dxdt ( u ( ϕ + t ϕ) + u ϕ( u + q)) dxdt (.4) for a.a. t ] T 1, T [ and for all nonnegative functions ϕ C 0 (R 3+1 ), vanishing in the neighborhood of the parabolic boundary Q ω {t = T 1 } ω [ T 1, T ]. The main result of the theory of suitable weak solutions, which we are going to use, is as follows. Lemma. There exist absolute positive constants ε 0 and c 0k, k = 1,,..., with the following property. Assume that the pair U and P is suitable weak solution to the Navier-Stokes equations in Q and satisfies the condition ( ) U 3 + P 3 dz < ε 0. (.5) Q Then, for any natural number k, k 1 U is Hölder continuous in Q( 1 ) and the following bound is valid: To formulate Lemma., we exploit the notation: max k 1 U(z) < c 0k. (.6) z Q( 1 ) z = (x, t), z 0 = (x 0, t 0 ); B(x 0, R) = { x x 0 < R}; Q(z 0, R) = B(x 0, R) ]t 0 R, t 0 [; B(r) = B(0, r), Q(r) = Q(0, r), B = B(1), Q = Q(1). Remark.3 For k = 1, Lemma. was proved essentially in [], see Corollary 1. For alternative approach, we refer the reader to [17], see Lemma 3.1. Cases k > 1 were treated in [8], see Proposition.1, with the help of the case k = 1 and regularity results for linear Stokes type systems. In fact, for the case k = 1, Lemma. is a consequence of scaling and the following statement. 7

Proposition.4 Given numbers θ ]0, 1/[ and M > 3, there exist two positive constants ε 1 (θ, M) and c 1 (M) such that, for any suitable weak solution v and p to the Navier-Stokes equations in Q, satisfying the additional conditions (v),1 < M, Y 1 (v, p) < ε 1, (.7) the following estimate is valid: Here and in what follows, we use the notation: (v) z0,r = Y θ (v, p) c 1 θ 3 Y1 (v, p). (.8) Y (z 0, R; v, p) = Y 1 (z 0, R; v) + Y (z 0, R; p), ( Y 1 1 (z 0, R; v) = Q(R) ( Y 1 (z 0, R; p) = R Q(R) 1 Q(R) Q(z 0,R) Q(z 0,R) Q(z 0,R) v dz, [p] x0,r = v (v) z0,r 3 dz ) 1 3, p [p] z0,r 3 dz ) 3, 1 B(R) B(x 0,R) Y 1 θ (v) = Y 1 (0, θ; v), Y θ (p) = Y (0, θ; p), p dx, Y θ (v, p) = Y (0, θ; v, p), (v),θ = (v) 0,θ, [p],θ = [p] 0,θ. Proof of Proposition.4 Assume that the statement of the proposition is false. This means that a number θ ]0, 1/[ and a sequence of suitable weak solutions v k and p k (in Q) exist such that: as k +, Y 1 (v k, p k ) = ε 1k 0 (.9) Y θ (v k, p k ) > c 1 ε 1k θ 3. (.10) The constant c 1 will be chosen later in order to get a contradiction. introduce new functions u k = (v k (v k ),1 )/ε 1k, q k = (p k [p k ],1 )/ε 1k. We 8

They satisfy the following relations and the system Y 1 (u k, q k ) = 1, (.11) Y θ (u k, q k ) > c 1 θ 3, (.1) t u k + 1 ε 1k div ((v k ),1 + ε 1k u k ) ((v k ),1 + ε 1k u k ) u k = q k, div u k = 0 in the sense of distributions. Without loss of generality, we may assume that: and } in Q (.13) u k u in L 3 (Q) q k q in L 3 (Q) (.14) (v k ),1 b in R 3 t u + div u b u = q div u = 0 in the sense of distributions. By (.11) and (.14), we have } in Q (.15) b < M, Y 1 (u, q) 1, [q(, t)],1 = 0 for all t ] 1, 0[. (.16) From the regularity theory for solutions to the Stokes system, see, for instance, [38], and from (.15), (.16), it follows that the function u is Hölder continuous in Q(3/4) and the following estimate is valid: Y 1 θ (u) c 1 (M)θ 3. (.17) On the other hand, choosing a cut-off function ϕ in an appropriate way in the local energy inequality, we find u k,,q(3/4) + u k,q(3/4) c 3 (M). (.18) Using the known multiplicative inequality, we derive from (.18) another estimate u k 10 3,Q(3/4) c 4(M). (.19) 9

It remains to make use of system (.13) and the inequalities in (.16). As a result, we have t u k L 3 ( (3/4),0;( c 5(M). (.0) W (B(3/4))) ) By well-known compactness arguments, we select a subsequence with the property u k u in L 3 (Q(3/4)). (.1) Now, taking into account (.1) and (.17), we pass to the limit in (.1) and find c 1 θ 3 c1 θ 3 + lim sup Yθ (q k ). (.) k To take the limit of the last term in the right hand side of (.), we decompose the pressure q k so that (see [35, 36, 37]) q k = q k 1 + q k, (.3) where the function q1 k is defined as a unique solution to the following boundary value problem: find q1(, k t) L 3 (B) such that q1(x, k t) ψ(x) dx = ε 1k u k (x, t) u k (x, t) : ψ(x) dx B B for all smooth test functions ψ subjected to the boundary condition ψ B = 0. It is easy to see that q k (, t) = 0 in B (.4) and, by the coercive estimates for Laplace s operator, we have the bound for q1: k q1(x, k t) 3 dx c6 ε 3 1k u k (x, t) 3 dx. (.5) B Here, c 6 is an absolute positive constant. Passing to the limit in (.), we show with the help of (.5) B c 1 θ 3 c1 θ 3 + lim sup Yθ (q). k (.6) k 10

By Poincare s inequality, (.6) can be reduced to the form ( c 1 θ 3 c1 θ 1 ) 3 + c7 θ lim sup q k k Q(θ) 3 3 dz. (.7) Q(θ) Since the function q(, k t) is harmonic in B, we have the estimate sup q(x, k t) 3 c8 q(x, k t) 3 dx x B(3/4) B and therefore 1 Q(θ) Q(θ) c 9 q k 3 c 9 dz θ ( 1 θ + 1 θ Q Q q k 1 3 dz ). q k 3 dz The latter inequality together with (.5) allows us to take the limit in (.7). As a result, we have c 1 θ 3 c1 θ 3 + c7 (c 9) 3 θ 3. (.8) If, from the very beginning, c 1 is chosen so that c 1 = ( c 1 + c 7 (c 9) 3 ), we arrive at the contradiction. Proposition.4 is proved. Proposition.4 admits the following iterations. Proposition.5 Given numbers M > 3 and β [0, /3[, we choose θ ]0, 1/[ so that c 1 (M)θ 3β 6 < 1. (.9) Let ε 1 (θ, M) = min{ε 1 (θ, M), θ 5 M/}. If then, for any k = 1,,..., (v),1 < M, Y 1 (v, p) < ε 1, (.30) θ k 1 (v),θ k 1 < M, Y θ k 1(v, p) < ε 1 ε 1, Y θ k(v, p) θ +3β 6 Y θ k 1(v, p). (.31) 11

Proof We use induction on k. For k = 1, this is nothing but Proposition.4. Assume now that statements (.31) are valid for s = 1,,..., k. Our goal is prove that they are valid for s = k+1 as well. Obviously, by induction, Y θ k(v, p) < ε 1 ε 1, and (v k ),1 = θ k (v),θ k θ k (v),θ k (v),θ k 1 + θ k (v),θ k 1 1 θ 5 Y θ k 1(v, p) + 1 θk 1 (v),θ k 1 < 1 θ 5 ε 1 + M/ M. Now, we make natural scaling: v k (y, s) = θ k v(θ k y, θ k s), p k (y, s) = θ k p(θ k y, θ k s) for (y, s) Q. We observe that v k and p k form suitable weak solution in Q. Since Y 1 (v k, p k ) = θ k Y θ k(v, p) < ε 1 ε 1 and we conclude (v k ),1 = θ k (v),θ k < M, Y θ (v k, p k ) c 1 θ 3 Y1 (v k, p k ) < θ +3β 6 Y 1 (v k, p k ), which is equivalent to the third relation in (.31). Proposition.5 is proved. A direct consequence of Proposition.5 and the scaling v R (y, s) = Rv(x 0 + Ry, t 0 + R s), p R (y, s) = R p(x 0 + Ry, t 0 + R s) is the following statement. Proposition.6. Let M, β, θ, and ε 1 be as in Proposition.5. Let a pair v and p be an arbitrary suitable weak solution to the Navier-Stokes equations in the parabolic cylinder Q(z 0, R), satisfying the additional conditions Then, for any k = 1,,..., we have R (v) z0,r < M, RY (z 0, R; v, p) < ε 1. (.3) Y (z 0, θ k R; v, p) θ +3β 6 k Y (z 0, R; v, p). (.33) 1

Proof of Lemma. We start with the case k = 1. We let ( ) A = U 3 + P 3 dz. Q Then, let M = 00, β = 1/3, and θ is chosen according to (.9) and fix. First, we observe that Q(z 0, 1/4) Q if z 0 Q(3/4) and 1 4 Y (z 0, 1/4; U, P ) c 10 (A 1 1 3 + A 3 ), 4 (U) z 0, 1 c 10 A 1 3 4 for an absolute positive constant c 10. Let us choose ε 0 so that Then, by (.5), we have c 10 (ε 1 3 0 + ε 3 0 ) < ε 1, c 10 ε 1 3 0 < 00. 1 4 Y (z 0, 1/4; U, P ) < ε 1, and thus, by Proposition.6, 1 4 (U) z 0, 1 < M, 4 Y (z 0, θ k /4; U, P ) θ k Y (z0, 1/4; U, P ) θ k ε1 for all z 0 Q(3/4) and for all k = 1,,... Hölder continuity of v on the set Q(/3) follows from Campanato s condition. Moreover, the quantity sup v(z) z Q(/3) is bounded by an absolute constant. The case k > 1 is treated with the help of the regularity theory for the Stokes equations and bootstrap arguments, for details, see [8], Proposition.1. Lemma. is proved. 13

3 Proof of the main results We start with the proof of Theorem 1.3, assuming that the statement of Theorem 1.4 is valid. Our approach is based on the reduction of the regularity problem to some problems from the theory of unique continuation and backward uniqueness for the heat operator. We follow the paper [40]. Proof of Theorem 1.3 The first observation is a consequence of (1.13) and can be formulated as follows t v(x, t) w(x) dx is continuous in [0, T ] for all w L 3. (3.1) R 3 This means that v(, t) 3 is bounded for each t [0, T ]. Using known Ladyzhenskaya s arguments, involving the coercive estimates (see [11, 4, 44]) and the uniqueness theorem for Stokes problem (see [14]), we can introduce the so-called associated pressure p and, since div v v L 4 (Q T ), we find 3 v L 4 (Q T ), t v, v, p L 4 (Q δ1,t ) (3.) 3 for any δ 1 > 0, where Q δ1,t = R 3 ]δ 1, T [. The pair v and p satisfies the Navier-Stokes equations a.e. in Q T. Moreover, by the pressure equation p = div div v v, (3.3) we have p L 3, (Q T ). (3.4) The pair v and p is clearly a suitable weak solution in any bounded cylinders of Q T. Moreover, by (3.), the local energy inequality holds as the identity. So, we can apply Theorem 1.4 and state that: for any z 0 R 3 ]0, T ], there exists a neighborhood O z0 of z 0 such that v is Hölder continuous in R 3 ]0, T ] O z0. (3.5) Indeed, for any z 0 R 3 ]0, T ] and for any R > 0 such that Q(z 0, R) Q T, making obvious scaling ṽ(x, t) = Rv(x 0 +Rx, t 0 +R t) and p(x, t) = R p(x 0 + Rx, t 0 + R t), we see that the pair ṽ and p satisfies all conditions of Theorem 14

1.4. This means that ṽ is Hölder continuous in Q(1/) and therefore v is Hölder continuous in Q(z 0, R/). So, (3.5) is a consequence of Theorem 1.4. Now, we are going to explain that, in turn, (3.5) implies Theorem 1.3. To this end, we note lim ( v 3 + p 3 ) dz = 0, Q(z0, R) Q T. z 0 + Q(z 0,R) Therefore, using scaling arguments, Lemma. and statement (3.5), we observe that max z R 3 [δ,t ] v(z) < C 1 (δ) < + (3.6) for all δ > 0. Setting w = v 3, we find from (1.13) and (3.6) w L, (Q T ) L (δ, T ; W 1 (R 3 )) and then, by the multiplicative inequality we deduce w(, t) 10 3 C w(, t) 5 w(, t) 3 5, (3.7) w L 10 (Q δ,t ) v L 5 (Q δ,t ) 3 for any δ > 0. On the other hand, since a L 3 J ( this is the necessary condition following from (3.1)), we apply Theorem 7.4 and conclude that v L 5 (Q δ0 ) for some δ 0 > 0. So, we have shown that Theorem 1.3 follows from (3.5). Theorem 1.3 is proved. Proof of Theorem 1.4 First, we note that v and p, satisfying conditions (1.15) and (1.16), form a suitable weak solution to the Navier-Stokes equations in Q. This can be verified with the help of usual mollification and the fact v L 4 (Q). The latter is just a consequence of the known multiplicative inequality. Second, we are going to prove two facts: t v(x, t) w(x) dx is continuous in [ (3/4), 0] B(3/4) for any w L 3 (B(3/4)) 15 (3.8)

and therefore sup v(, t) 3,B(3/4) v 3,,Q. (3.9) (3/4) t 0 We can justify (3.8) as follows. Using the local energy inequality, we can find the bound for v,q(5/6) via v 3,,Q and p 3,Q only. Then, by the known multiplicative inequality, we estimate the norm v 4,Q(5/6). Hence, div v v L 4 (Q(5/6)). Now, using a suitable cut-off function, the L s,l - 3 coercive estimates for solutions to the non-stationary Stokes system, known duality arguments, we find the following bound Q(3/4) ( ) v 4 + t v 4 3 + v 4 4 3 + p 3 dz c 01, (3.10) with a constant c 01 depending on the norms v 3,,Q and p 3,Q only. In particular, it follows from (3.10) that v C([ (3/4), 0]; L 4 (B(3/4))) which, 3 in turn, implies (3.8). As in the proof of Proposition.4, we can present the pressure p in the form p = p 1 + p, where the function p 1 (, t) is a unique solution to the following boundary value problem: find p 1 (, t) L 3 (B) such that p 1 (x, t) ψ(x) dx = v(x, t) v(x, t) : ψ(x) dx B B for all smooth functions ψ satisfying the boundary condition ψ B = 0. Then, p (, t) = 0 in B. The same arguments, as in Section, lead to the estimates and ( 0 p, 3,B(3/4) ] 1,0[ = p 1 3,,Q c 1 v 3,,Q (3.11) 1 ) sup p (x, t) 3 3 dt x B(3/4) c 1 ( p 3,Q + v 3,,Q ), where c 1 is an absolute positive constant. (3.1) 16

Assume that the statement of Theorem 1.4 is false. Let z 0 Q(1/) be a singular point, see the definition of regular points in the proof of Theorem 1.3. Then, as it was shown in [39], there exists a sequence of positive numbers R k such that R k 0 as k + and A(R k ) sup t 0 R k t t 0 1 R k B(x 0,R k ) v(x, t) dx > ε (3.13) for all k N. Here, ε is an absolute positive constant. We extend functions v and p to the whole space R 3+1 by zero. Extended functions will be denoted by ṽ and p, respectively. Now, we let v R k (x, t) = R k ṽ(x 0 + R k x, t 0 + R kt), p R k (x, t) = R k p(x 0 + R k x, t 0 + R kt), p R k 1 (x, t) = R k p 1 (x 0 + R k x, t 0 + R kt), p R k (x, t) = R k p (x 0 + R k x, t 0 + R kt), where p 1 and p are extensions of p 1 and p, respectively. Obviously, for any t R, R 3 v Rk (x, t) 3 dx = R 3 ṽ(x, t 0 + Rkt) 3 dx, (3.14) R p k 1 (x, t) 3 dx = R 3 and, for any Ω R 3, sup p R k (x, t) 3 dt = Rk x Ω R R Hence, without loss of generality, one may assume that where divu = 0 in R 3 R and p R k R 3 p 1 (x, t 0 + Rkt) 3 dx (3.15) sup p (x 0 + R k x, s) 3 ds. (3.16) x Ω v R k u in L (R; L 3 ) as k +, (3.17) p R k 1 q in L (R; L 3 ) as k +, (3.18) 0 in L 3 (R; L (Ω)) as k + (3.19) 17

for any Ω R 3. For justification of (3.18) and (3.19), we take into account identities (3.15), (3.16) and bounds (3.11), (3.1). To extract more information about boundedness of various norms of functions v R k and p R k, let us fix a cut-off function φ C 0 (R 3+1 ) and introduce the function φ R k in the following way φ(y, τ) = R k φ R k (x 0 + R k y, t 0 + R kτ), y R 3 τ R. We choose R k so small to ensure suppφ {(y, τ) t 0 + R kτ ] (3/4), (3/4) [, x 0 + R k y B(3/4)} = suppφ R k B(3/4) ] (3/4), (3/4) [. Then, since the pair v and p is a suitable weak solution, we have 0 1 B φ R k v dz 0 1 B { } v ( φ R k + t φ R k ) + v φ R k( v + p) dz and after changing variables we arrived at the inequality { } φ v Rk dz v R k ( φ + t φ) + v Rk φ( v R k + p R k ) dz. R R 3 R R 3 Now, our goal is to estimate p R k 3,Ω ]a,b[ for all Ω R3 and for all < a < b < +. We find p R k 3,Ω ]a,b[ pr k 1 3,Ω ]a,b[ + pr k 3,Ω ]a,b[ [ ( b c (a, b, Ω) p R k 1 3,,R3 R + [ ( c (a, b, Ω) p R k 1 3,,R3 R + R k a R Ω ) sup p R k (y, t) 3 ] 3 dxdt y Ω ) sup p (x 0 + R k y, s) 3 ] 3 ds y Ω c (a, b, Ω)( p 3,Q + v 3,,Q). So, from the last two inequalities, we deduce the bound ) ( p R k 3 + v R k dz c 3 (Q) < + (3.0) Q 18

for any domain Q R 3+1 with a constant c 3 in (3.0) independent of R k. Then, we apply known arguments, including multiplicative inequalities, the L s,l -coercive estimates for solutions to the non-stationary Stokes equations, and duality. As a result, we find ( v R k 4 + t v R k 4 3 + v R k 4 3 + p R k 4 3 ) dz c4 (Q). (3.1) Q The latter together with (3.17) implies for Q R 3+1. Let us show that, in addition, v R k u in L 3 (Q) (3.) v R k u in C([a, b]; L (Ω)) (3.3) for any < a < b < + and for any Ω R 3. Indeed, by (3.1), v R k u in C([a, b]; L 4 (Ω)) 3 and then (3.3) can be easily derived from the interpolation inequality v R k (, t + t) v R k (, t),ω v R k (, t + t) v R k (, t) 5 4 3,Ω vr k (, t + t) v R k (, t) 3 5 3,Ω and from (3.17). Now, we combine all information about limit functions u and q, coming from (3.14) (3.3), and conclude that: ( u 4 + u + t u 4 3 + u 4 4 3 + q 3 ) dz c3 (Q) (3.4) Q for any Q R 3+1 ; for any < a < b < + and for any Ω R 3 ; u C([a, b]; L (Ω)) (3.5) functions u and q satisfy the Navier-Stokes equations a.e. in R 3+1 ; (3.6) 19

φ u dz = R R 3 R R 3 { } u ( φ + t φ) + u φ( u + q) dz (3.7) for all functions φ C0 (R 3+1 ). It is easy to show that, according to (3.4) (3.7), the pair u and q is a suitable weak solution to the Navier-Stokes equations in Ω [a, b] for any bounded domain Ω R 3 and for any < a < b < +. Moreover, according to (3.13), sup t 0 R k t t 0 1 R k B(x 0,R k ) v(x, t) dx = sup 1 t 0 B(0,1) v R k (x, t) dx > ε for all k N and, by (3.3), we find sup u(x, t) dx ε. (3.8) 1 t 0 B(0,1) Let us proceed the proof of Theorem 1.4. We are going to show that there exist some positive numbers R and T such that, for any k = 0, 1,..., the function k u is Hölder continuous and bounded on the set (R 3 \ B(R /)) ] T, 0]. To this end, let us fix an arbitrary number T > and note that Therefore, 0 4T ( u 3 3 + q )dz < +. R 3 0 ( u 3 + q 3 )dz 0 as R +. 4T R 3 \B(0,R) This means that there exists a number R (ε 0, T ) > 4 such that 0 ( u 3 + q 3 )dz < ε0. (3.9) 4T R 3 \B(0,R /4) 0

Now, assume that z 1 = (x 1, t 1 ) (R 3 \ B(R /)) ] T, 0]. Then, Q(z 1, 1) B(x 1, 1) ]t 1 1, t 1 [ (R 3 \ B(0, R /4)) ] 4T, 0]. So, by (3.9), t 1 ( u 3 + q 3 )dz < ε0 (3.30) t 1 1 B(x 1,1) for any z 1 (R 3 \ B(R /)) ] T, 0], where T > and R > 4. Then, it follows from (3.30) and from Lemma. that, for any k = 0, 1,..., max k u(z) c 0k < + (3.31) z Q(z 1,1/) and k u is Hölder continuous on (R 3 \ B(R /)) ] T, 0]. Now, let us introduce the vorticity ω of u, i.e., ω = u. The function ω meets the equation t ω + u k ω,k ω k u,k ω = 0 in (R 3 \ B(R )) ] T, 0]. Recalling (3.31), we see that, in the set (R 3 \ B(R )) ] T, 0], the function ω satisfies the following relations: for some constant M > 0 and Let us show that t ω ω M( ω + ω ) (3.3) ω c 00 + c 01 < +. (3.33) ω(x, 0) = 0, x R 3 \ B(R ). (3.34) To this end, we take into account the fact that u C([ T, 0]; L ) and find ( ) 1 u(x, 0) dx ( B(x,1) B(x,1) ) 1 ( v R k (x, 0) u(x, 0) dx + B(x,1) ) 1 v R k (x, 0) dx 1

( v R k u C([ T,0];L ) + B 1 6 v R k u C([ T,0];L ) + B 1 6 B(x,1) ( ) 1 v R k (x, 0) 3 3 dx v(y, t 0 ) 3 dy ) 1 3. B(x 0 +R k x,r k ) Since v(, t) 3,B(3/4) is bounded for any t [ (3/4), 0], see (3.9), we show that, by (3.3), u(x, 0) dx = 0 B(x,1) for all x R 3. So, (3.34) is proved. Relations (3.3) (3.34) allow us to apply the backward uniqueness theorem of Section 5, see Theorem 5.1, and conclude that If we show that ω(z) = 0 z (R 3 \ B(R )) ] T, 0]. (3.35) ω(, t) = 0 in R 3 (3.36) for a.a. t ] T, 0[, then we are done. Indeed, by (3.36), the function u(, t) is harmonic and has the finite L 3 -norm. It turn, this fact leads to the identity u(, t) = 0 for a.a. t ] T, 0[. This contradicts with (3.8). So, our goal is to show that (3.35) implies (3.36). To simplify our notation, we let T = T /, R = R. We know that functions u and q meet the equations: t u + div u u = q, div u = 0, u = 0, u = 0 (3.37) in the set (R 3 \ B(R/)) ] T, 0]. From (3.37), we deduce the following bound ( ) max k u(z) + k t u(z) + k q(z) c 1 0k < + (3.38) z Q 0 for all k = 0, 1,... Here, Q 0 = (R 3 \ B(R)) ] T, 0]. Next, we fix a smooth cut-off function ϕ C 0 (R 3 ) subjected to the conditions: ϕ(x) = 1 if x B(R), ϕ(x) = 0 if x / B(3R). Then, we let w = ϕu, r = ϕq. New functions w and r satisfy the system t w + div w w w + r = g div w = u ϕ (3.39)

in Q = B(4R) ] T, 0[ and where w B(4R) [ T,0] = 0, (3.40) g = (ϕ ϕ) div u u + u u ϕ + q ϕ u ϕ u ϕ. The function g satisfies the conditions: g(x, t) = 0 if x B(R) or x / B(3R), (3.41) ( ) sup k g(z) + k t g(z) c 0k < + (3.4) z Q 0 for all k = 0, 1,... Obviously, (3.4) follows from (3.31), (3.38), and (3.41). Unfortunately, the function w is not solenoidal. For this reason, we introduce functions w and r as a solution to the Stokes system: w + r = 0, div w = u ϕ in Q with the homogeneous boundary condition w B(4R) [ T,0] = 0. According to the regularity theory for stationary problems and by (3.38), we can state sup ( k t w(z) + k w(z) + k r(z) ) c 3 0k < + (3.43) z Q for all k = 0, 1,... Setting U = w w and P = r r, we observe that, by (3.41) and (3.4), U and P meet the Navier-Stokes system with linear lower order terms: t U + div U U U + P = div(u w in Q, (3.44) + w U) + G, div U = 0 U B(4R) [ T,0] = 0, (3.45) where G = div w w + g t w, and, taking into account (3.4) and (3.43), we have sup k G(z) c 4 0k < + (3.46) z Q 3

for all k = 0, 1,... Standard regularity results and the differential properties of u and q, described in (3.4), (3.5), and (3.6), lead to the following facts about smoothness of functions U and P : U L 3, (Q ) C([ T, 0]; L (B(4R))) L ( T, 0; W 1 (B(4R))), t U, U, P L 4 (Q ). 3 Let t 0 ] T, 0[ be chosen so that U(, t 0 ),B(4R) < +. (3.47) Then, by the short time unique solvability results for the Navier-Stokes system (see [14, 16]), we can find a number δ 0 > 0 such that t U, U, P L (B(4R) ]t 0, t 0 + δ 0 [). In turn, the regularity theory for linear systems implies the bounds sup t 0 +ε<t<t 0 +δ 0 ε sup k U(x, t) c 5 0k < + x B(4R) for all k = 0, 1,... and for some nonnegative number ε < δ 0 /4. They immediately imply information about smoothness of the original function u: sup t 0 +ε<t<t 0 +δ 0 ε sup k u(x, t) c 6 0k < + x B(4R) for all k = 0, 1,... Hence, we can state that t ω ω M( ω + ω ) and ω M 1 in B(4R) ]t 0 +ε, t 0 +δ 0 ε[ for some positive constants M and M 1. But we know that ω(z) = 0 if z (B(4R) \ B(R)) ]t 0 + ε, t 0 + δ 0 ε[. By the unique continuation theorem of Section 4, see Theorem 4.1, we conclude that: ω = 0 in B(4R) ]t 0 + ε, t 0 + δ 0 ε[. Since (3.47) holds for a.a. t 0 ] T, 0[, we find ω(, t) = 0 in R 3 for a.a. t ] T, 0[. Repeating the same arguments in the interval ] T, T /[, we arrive at (3.36). Theorem 1.3 is proved. We would like to note that the final part of the proof of Theorem 1.3 can be carried out in different ways. For example, we could argue as follows. We should expect that the function U(, t) and therefore the function u(, t) are analytic one s in the ball B(R) for t 0 + ε < t < t 0 + δ 0 ε, see [5]. This means that the vorticity ω is also an analytic function in space variables on the same set. Since ω = 0 outside B(R), we may conclude that ω = 0 in R 3 ]t 0 + ε, t 0 + δ 0 ε[ and so on. 4

4 Unique Continuation Through Spatial Boundaries In this section, we are going to discuss known facts from the theory of unique continuation for differential inequalities. We restrict ourselves to justification only of those statements which are going to be used in what follows and which can be easily reproved within our unified approach. We hope that this makes our paper more self-contained and more convenient for reading. For advanced theory in this direction, we refer the reader to the paper [4], see also the list of quotations there. We will work with the backward heat operator t + rather than the more usual heat operator t since this will save us writing some minus signs in many formulae. In the space-time cylinder Q(R, T ) B(R) ]0, T [ R 3 R 1, we consider a vector-valued function u = (u i ) = (u 1, u,..., u n ), satisfying three conditions: u W,1 (Q(R, T ); R n ); (4.1) t u + u c 1 ( u + u ) a.e. in Q(R, T ) (4.) for some positive constant c 1 ; u(x, t) C k ( x + t) k (4.3) for all k = 0, 1,..., for all (x, t) Q(R, T ), and for some positive constants C k. Here, W,1 (Q(R, T ); R n ) { u + u + u + t u L (Q(R, T ))}. Condition (4.3) means that the origin is zero of infinite order for the function u. Theorem 4.1 Assume that a function u satisfies conditions (4.1) (4.3). Then, u(x, 0) = 0 for all x B(R). Remark 4. For more general results in this direction, we refer the reader to the paper [4] of Escauriaza-Fernández. Without loss of generality, we may assume that T 1. Theorem 4.1 is an easy consequence of the following lemma. 5

Lemma 4.3 Suppose that all conditions of Theorem 4.1 hold. Then, there exist a constant γ = γ(c 1 ) ]0, 3/16[ and absolute constants β 1 and β such that u(x, t) c (c 1, n)a 0 (R, T )e x 4t (4.4) for all (x, t) Q(R, T ) satisfying the following restrictions: 0 < t γt, x β 1 R, β t x. Here, A 0 max u(x, t) + T u(x, t). (x,t) Q( 3 4 R, 3 4 T ) Remark 4.4 According to the statement of Lemma 4.3, u(x, 0) = 0 if x β 1 R. Remark 4.5 From the regularity theory for parabolic equations (see [18]), it follows that ( ) 1 A 0 c 3 (R, T ) u dz. Q(R,T ) Proof of Lemma 4.3 We let λ = t and ϱ = x /λ. Suppose that x 3R and 8t 8 x. Then, as it is easy to verify, we have ϱ 4 and λy B(3R/4) if y B(ϱ); λ s ]0, 3/4[ if s ]0, [ under the condition 0 < γ 3/16. Thus the function v(y, s) = u(λy, λ s) is well defined on Q(ϱ, ) = B(ϱ) ]0, [. This function satisfies the conditions: in Q(ϱ, ); s v + v c 1 λ( v + v ) (4.5) v(y, s) C k( y + s) k (4.6) for all k = 0, 1,... and for all (y, s) Q(ϱ, ). Here, C k = C kλ k. Given ε > 0, we introduce two smooth cut-off functions with the properties: { 1, (y, s) Q(ϱ 1, 3/) 0 ϕ(y, s) = 1, 0, (y, s) / Q(ϱ, ) { 1, s ]ε, [ 0 ϕ ε (s) = 1. 0, s ]0, ε[ 6

We let w = ϕv and w ε = ϕ ε w. Obviously, (4.5) implies the following inequality: s w ε + w ε c 1 λ( w ε + w ε ) +c 4 ( ϕ v + ϕ v + ϕ v + s ϕ v ) + c 4 ϕ ε v. (4.7) The crucial point is the application of the following Carleman-type inequality, see Section 6 for details, Proposition 6.1, to the function w ε Q(ϱ,) c 5 Q(ϱ,) h a (s)e y 4s ( w ε + w ε ) dyds h a (s)e y 4s s w ε + w ε dyds. (4.8) Here, c 5 is an absolute positive constant, a is an arbitrary positive number, and h(t) = te 1 t 3. We let A = max (y,s) Q(ϱ,)\Q(ϱ 1, 3 ) v(y, s) + v(y, s) and choose γ sufficiently small in order to provide the condition 10c 5 c 1λ 0c 5 c 1γ < 1. (4.9) Condition (4.9) makes it possible to hide the strongest term in the right hand side of (4.8) into the left hand side of (4.8). So, we derive from (4.7) (4.9) the following relation Q(ϱ,) h a (s)e y 4s ( w ε + w ε ) dyds c 6 A Q(ϱ,) h a (s)e y 4s χ(y, s) dyds (4.10) +c 6 1 ε Q(ϱ,ε) h a (s)e y 4s v dyds. Here, χ is the characteristic function of the set Q(ϱ, ) \ Q(ϱ 1, 3/). We fix a and take into account (4.6). As a result of the passage to the limit as 7

ε 0, we find from (4.10) D Q(ϱ 1,3/) h a (s)e y 4s ( v + v ) dyds c 6 A Q(ϱ,) h a (s)e y 4s χ(y, s) dyds (4.11) ( c 6A h a (3/) + ρ n 1 0 h a (s)e (ϱ 1) 4s ) ds. Since ϱ 4, it follows from (4.11) that: ( D c 7 A h a (3/) + ρ n 1 0 h a (s)e ϱ 8s ds ). (4.1) In (4.1), the constant c 7 depends on n and c 1 only. Given positive number β, we can take a number a in the following way a = βϱ ln h(3/). (4.13) This is legal, since h(3/) > 1. Hence, by (4.13), inequality (4.1) can be reduced to the form D c 7 A e βρ( 1 + ρ n 1 e βϱ 0 h a (s)e βϱ ϱ 8s ds ). We fix β ]0, 1/64[, say, β = 1/100. Then, the latter relation implies the estimate D c 7(c 1, n)a e βϱ( 1 + 0 h a (s)e ϱ 16s ds ). (4.14) It is easy to check that β ln(3/) and therefore g (s) 0 if s ]0, [, where 1 g(s) = h a (s)e ϱ 16s and a and ϱ satisfy condition (4.13). So, we have D c 8 (c 1, n)a e βϱ, (4.15) 8

where β is an absolute positive constant. By the choice of ϱ and λ, we have B(µ x, 1) B(ϱ 1) for any µ ]0, 1]. λ Then, setting Q = B(µ x, 1) ]1/, 1[, we find λ Q D Q e y v dyds. (4.16) Observing that y µ x + if y B(µ x, 1) and letting µ = β, we λ λ derive from (4.15) and (4.16) the following bound v dyds c 8A µ ( β+ e ) x t = c 8A x β e t. (4.17) On the other hand, the regularity theory for linear backward parabolic equations give us: v(µx/λ, 1/) c 9 (c 1, n) v dyds. (4.18) Combining (4.17) and (4.18), we show Q u( βx, t) = u(µx, t) = v(µx/λ, 1/) c 9A x β e t. Changing variables x = βx, we have u( x, t) c 9Ae x 4t for x β 1 R and x β t with β 1 = 3/8 β and β = 16β. It remains to note that λ T and A Lemma 4.3 is proved. max u(x, t) + λ u(x, t). (x,t) Q( 3 4 R, 3 4 T ) 5 Backward Uniqueness for Heat Operator in Half Space In this section, we deal with a backward uniqueness problem for the heat operator. Our approach is due to [7], see also [5] and [6]. 9

Let R n + = {x = (x i ) R n x n > 0} and Q + = R n + ]0, 1[. We consider a vector-valued function u : Q + R n, which is sufficiently regular and satisfies t u + u c 1 ( u + u ) in Q + (5.1) for some c 1 > 0 and u(, 0) = 0 in R n +. (5.) Do (5.1) and (5.) imply u 0 in Q +? We prove that the answer is positive if we impose natural restrictions on the growth of the function u at infinity. For example, we can consider u(x, t) e M x (5.3) for all (x, t) Q + and for some M > 0. Natural regularity assumptions, under which (5.1) (5.3) can be considered are, for example, as follows: } u and distributional derivatives t u, u are square (5.4) integrable over bounded subdomains of Q +. We can formulate the main result result of this section. Theorem 5.1 Using the notation introduced above, assume that u satisfies conditions (5.1) (5.4). Then u 0 in Q +. This extends the main result of [5] and [6], where an analogue of Theorem 5.1 was proved for Q + replaced with (R n \ B(R)) ]0, T [. Similarly to those papers, the proof of Theorem 5.1 is based on two Carleman-type inequalities, see (6.1) and (6.1). Such results are of interest in control theory, see for example [7]. The point is that the boundary conditions are not controlled by our assumptions. It is an easy exercise for the reader to prove that Theorem 5.1 is true for functions u : Q + R m with 1 m < +. We start with proofs of several lemmas. The first of them plays the crucial role in our approach. It enables us to apply powerful technique of Carleman s inequalities. Lemma 5. Suppose that conditions (5.1), (5.), and (5.4) are fulfilled. There exists an absolute positive constant A 0 < 1/3 with the following properties. If u(x, t) e A x (5.5) 30

for all (x, t) Q + and for some A [0, A 0 ], then there are constants β(a) > 0, γ(c 1 ) ]0, 1/1[, and c (c 1, A) > 0 such that for all (x, t) (R n + + e n ) ]0, γ[. u(x, t) c e 4A x e β x n t (5.6) Proof In what follows, we always assume that the function u is extended by zero to negative values of t. According to the regularity theory of solutions to parabolic equations, see [18], we may assume u(x, t) + u(x, t) c 3 e A x (5.7) for all (x, t) (R n + + e n ) ]0, 1/[. We fix x n > and t ]0, γ[ and introduce the new function v by usual parabolic scaling v(y, s) = u(x + λy, λ s t/). The function v is well defined on the set Q ρ = B(ρ) ]0, [, where ρ = (x n 1)/λ and λ = 3t ]0, 1/[. Then, relations (5.1), (5.), and (5.7) take the form: s v + v c 1 λ( v + v ) a.e. in Q ρ ; (5.8) for (y, s) Q ρ ; v(y, s) + v(y, s) c 3 e 4A x e 4Aλ y (5.9) v(y, s) = 0 (5.10) for y B(ρ) and for s ]0, 1/6]. In order to apply inequality (6.1), we choose two smooth cut-off functions: { 0 y > ρ 1/ φ ρ (y) = 1 y < ρ 1, φ t (s) = { 0 7/4 < s < 1 0 < s < 3/. These functions take values in [0, 1]. In addition, function φ ρ satisfies the conditions: k φ ρ < C k, k = 1,. We let η(y, s) = φ ρ (y)φ t (s) and w = ηv. It follows from (5.8) that s w + w c 1 λ( w + w ) + χc 4 ( v + v ). (5.11) 31

Here, c 4 is a positive constant depending on c 1 and C k only, χ(y, s) = 1 if (y, s) ω = {ρ 1 < y < ρ, 0 < s < } { y < ρ 1, 3/ < s < } and χ(y, s) = 0 if (y, s) / ω. Obviously, function w has the compact support in R n ]0, [ and we may use inequality (6.1), see Proposition 6.1. As a result, we have I Q ρ h a (s)e y 4s ( w + w ) dyds c 0 10(c 1λ I + c 6I 1 ), (5.1) where I 1 = χ(y, s)h a (s)e y 4s ( v + v ) dyds. Q ρ Choosing γ = γ(c 1 ) sufficiently small, we can assume that the inequality c 0 10c 1λ 1/ holds and then (5.1) implies On the other hand, if A < 1/3, then I c 5 (c 1 )I 1. (5.13) 8Aλ 1 4s < 1 8s (5.14) for s ]0, ]. By (5.9) and (5.14), we have I 1 c 3e 8A x 0 B(ρ) c 6 e 8A x [h a (3/) + χ(y, s)h a (s)e y 8s 0 h a (s)e (ρ 1) 8s dyds ] ds. (5.15) Now, taking into account (5.15), we deduce the bound D 1 w dyds = 1 v dyds c 7 B(1) Q ρ 1 B(1) h a (s)e y 4s ( w + w ) dyds 1 3

c 8 (c 1 )e 8A x [ h a (3/) + = c 8 e 8A x βρ [ h a (3/)e βρ + 0 We can take β = 8A < 1/56 and then choose a = βρ / ln h(3/). Since ρ x n, such a choice leads to the estimate D c 8 e 8A x e βρ[ 1 + 0 h a (s)e ρ 3s ds ] 0 h a (s)e βρ ρ 3s ds ]. ] g(s) ds, where g(s) = h a (s)e ρ 64s. It is easy to check that g (s) 0 for s ]0, [ if β < 1 ln h(3/). So, we have 96 D c 8 e 8A x e βρ c 8 e 8A x e βx n 1t. (5.16) On the other hand, the regularity theory implies v(0, 1/) = u(x, t) c 8D. (5.17) Combining (5.16) and (5.17), we complete the proof of Lemma 5.. Lemma 5. is proved. Next lemma will be a consequence of Lemma 5. and the second Carleman inequality (see (6.1)). Lemma 5.3 Suppose that the function u satisfies conditions (5.1), (5.), (5.4), and (5.5). There exists a number γ 1 (c 1, c ) ]0, γ/] such that for all x R n + and for all t ]0, γ 1 [. u(x, t) = 0 (5.18) 33

Proof As usual, by Lemma 5. and by the regularity theory, we may assume u(x, t) + u(x, t) c 9 (c 1, A)e 8A x e β x n t (5.19) for all x R n + + 3e n and for all t ]0, γ/]. By scaling, we define function v(y, s) = u(λy, λ s γ 1 ) for (y, s) Q + with λ = γ 1. This function satisfies the relations: s v + v c 1 λ( v + v ) a.e. in Q + ; (5.0) for all y R n + and for all s ]0, 1/[; v(y, s) = 0 (5.1) βλ y n v(y, s) + v(y, s) c 9 e 8Aλ y e (λ s γ 1) c 9 e 8Aλ y e β y n s (5.) for all 1/ < s < 1 and for all y R n + + 3 λ e n. Since A < 1/3 and λ γ 1/ 1, (5.) can be reduced to the form v(y, s) + v(y, s) c 11 e y 48 e β y n s (5.3) for the same y and s as in (5.). Let us fix two smooth cut-off functions: { 0 yn < 3 ψ 1 (y n ) = + 1 λ 1 y n > 3 + 3 λ, and ψ (r) = { 1 r > 1/ 0 r < 3/4. We set (see Proposition 6. for the definition of φ (1) and φ () ) φ B (y n, s) = 1 a φ() (y n, s) B = (1 s) yα n s B, α where α ]1/, 1[ is fixed, B = a φ() ( 3 +, 1/), and λ η(y n, s) = ψ 1 (y n )ψ (φ B (y n, s)/b), w(y, s) = η(y n, s)v(y, s). 34

Although function w is not compactly supported in Q 1 + = (R 3 + + e n ) ]0, 1[, but, by the statement of Lemma 5. and by the special structure of the weight in (6.1), we can claim validity of (6.1) for w. As a result, we have s e φ(1) e aφ B ( w + w ) dyds Q 1 + c s e φ(1) e aφ B s w + w dyds. Q 1 + Arguing as in the proof of Lemma 5., we can select γ 1 (c 1, c ) so small that I s e aφ B ( w + w )e y 4s dyds Q 1 + c 10 (c 1, c ) χ(y n, s)(sy n ) e aφ B ( v + v )e y 4s (R n + +( 3 λ +1)en) ]1/,1[ where χ(y n, s) = 1 if (y n, s) ω, χ(y n, s) = 0 if (y n, s) / ω, and dyds, ω {(y n, s) y n > 1, 1/ < s < 1, φ B (y n, s) < D/}, where D = φ B ( 3 + 3, 1 ) > 0. Now, we wish to estimate the right hand λ side of the last inequality with the help of (5.3). We find I c 11 e Da + 1 3 λ +1 1/ (y n s) e β y n s dyn ds R n 1 e ( 1 4 1 4s ) y dy. Passing to the limit as a +, we see that v(y, s) = 0 if 1/ s < 1 and φ B (y n, s) > 0. Using unique continuation through spatial boundaries, see Section 4, we show that v(y, s) = 0 if y R n + and 0 < s < 1. Lemma 5.3 is proved. Now, Theorem 5.1 follows from Lemmas 5. and 5.3 with the help of more or less standard arguments. We shall demonstrate them just for completeness. Lemma 5.4 Suppose that the function u meets all conditions of Lemma 5.3. Then u 0 in Q +. 35

Proof By Lemma 5.3, u(x, t) = 0 for x R n + and for t ]0, γ 1 [. By scaling, we introduce the function u (1) (y, s) = u( 1 γ 1 y, (1 γ 1 )s + γ 1 ). It easy to check that function u (1) is well-defined in Q + and satisfies all conditions of Lemma 5.3 with the same constants c 1 and A. Therefore, u (1) (y, s) = 0 for y n > 0 and for 0 < s < γ 1. The latter means that u(x, t) = 0 for x n > 0 and for 0 < t < γ = γ 1 + (1 γ 1 )γ 1. Then, we introduce the function u () (y, s) = u( 1 γ y, (1 γ )s + γ ), (y, s) Q +, and apply Lemma 5.3. After k steps we shall see that u(x, t) = 0 for x n > 0 and for 0 < t < γ k+1, where γ k+1 = γ k + (1 γ k )γ 1 1. Lemma 5.4 is proved. Proof of Theorem 5.1 Assume that A 0 < M. Then λ A 0 < M 1. Introducing function v(y, s) = u(λy, λ s), (y, s) Q +, we see that this function satisfies all conditions of Lemma 5.4 with constants c 1 and A = 1A 0. Therefore, u(x, t) = 0 for x n > 0 and for 0 < t < A 0. Now, we repeat M arguments of Lemma 5.4, replacing γ 1 to A 0 and A to M, and end up with M the proof of the theorem. Theorem 5.1 is proved. 6 Carleman-Type Inequalities The first Carleman-type inequality is essentially the same as the one used in [5] and [6] (see also [3], [8], and [46]). Proposition 6.1 For any function u C 0 (R n ]0, [; R n ) and for any positive number a, the following inequality is valid: R n ]0,[ c 0 h a (t)e x 4t R n ]0,[ ( a t u + u ) dxdt h a (t)e x 4t t u + u dxdt. Here, c 0 is an absolute positive constant and h(t) = te 1 t 3. (6.1) Proof of Proposition 6.1 Our proof follows standard techniques used in the L -theory of Carleman inequalities, see for example [13] and [46]. 36

x 8t Let u be an arbitrary function from C0 (R n ]0, [; R n ). We set φ(x, t) = (a + 1) ln h(t) and v = e φ u. Then, we have Lv e φ ( t u + u) = t v div(v φ) v φ + v + ( φ t φ)v. The main trick in the above approach is the decomposition of operator tl into symmetric and skew symmetric parts, i.e., tl = S + A, (6.) where and Sv t( v + ( φ t φ)v) 1 v (6.3) Av 1 ( t(tv) + t t v) t(div(v φ) + v φ). (6.4) Obviously, t e φ t u + u dxdt = t Lv dxdt = Sv dxdt + Av dxdt + [S, A]v v dxdt, (6.5) where [S, A] = SA AS is the commutator of S and A. Simple calculations show that I [S, A]v v dxdt = = 4 t [ φ,ij v,i v,j + φ,ij φ,i φ,j v ] dxdt + t v ( t φ t φ φ) dxdt + t v dxdt t v ( φ t φ) dxdt. (6.6) Here and in what follows, we adopt the convention on summation over repeated Latin indices, running from 1 to n. Partial derivatives in spatial variables are denoted by comma in lower indices, i.e., v,i = v x i, v = (v i,j ), etc. Given choice of function φ, we have [ ( h I = (a + 1) t (t) ) h (t)) ] v dxdt = a + 1 t v dxdt. (6.7) h(t) th(t) 3 37

By the simple identity v = 1 ( t + ) v v ( t v + v), (6.8) we find t v dxdt = t v dxdt t v Lv dxdt In our case, + t v ( φ t φ) dxdt. φ t φ = φ + (a + 1) h (t) h(t). The latter relation (together with (6.7)) implies the bound t ( v + v φ ) dxdt 3I t v Lv dxdt b 1 t Lv dxdt, (6.9) (6.10) where b 1 is an absolute positive constant. Since e φ u v + v φ, (6.11) it follows from (6.5) (6.10) that ( ) h a (t)(th 1 (t)) (a + 1) u + u e x 4t dxdt t b h a (t)(th 1 (t)) t u + u e x 4t dxdt. Here, b is an absolute positive constant. Inequality (6.1) is proved. The second Carleman-type inequality is, in a sense, an anisotropic one. Proposition 6. Let φ = φ (1) + φ (), where φ (1) (x, t) = x and φ () (x, t) = a(1 t) xα n, x = (x 8t t α 1, x,..., x n 1 ) so that x = (x, x n ), and e n = (0, 0,..., 0, 1). Then, for any function u 38

C0 ((R n + + e n ) ]0, 1[; R n ) and for any number a > a 0 (α), the following inequality is valid: ( ) t e φ(x,t) a u + u dxdt t t (R n + +e n) ]0,1[ c (R n + +e n) ]0,1[ t e φ(x,t) t u + u dxdt. Here, c = c (α) is a positive constant and α ]1/, 1[ is fixed. (6.1) Proof Let u C 0 (Q 1 +; R n ). We are going to use formulae (6.) (6.6) for new functions u, v, and φ. All integrals in those formulae are taken now over Q 1 +. First, we observe that Therefore, φ (1) (x, t) = x 4t, φ = φ (1) + φ () φ() (x, t) = αa 1 t t α x α 1 n e n. (6.13) φ (1) φ () = 0, φ = φ (1) + φ (). (6.14) Moreover, φ (),ij = φ (1),ij = φ = φ (1) + φ (), δ ij 4t if 1 i, j n 1 0 if i = n or j = n 0 if i n or j n α(α 1)a 1 t t α x α n if i = n and j = n,. (6.15) In particular, (6.15) implies φ,ij φ,i φ,j = 1 4t φ(1) + α(α 1)a 1 t x α t α n φ () 1 4t x 16t 3. (6.16) 39

Using (6.14) (6.16), we present integral I in (6.6) in the following way: I = I 1 + I + t v dxdt, (6.17) where I s = 4 t [ φ (s),ij v,i v,j + φ (s),ij φ(s),i φ(s),j v ] dxdt + t v ( t φ (s) t φ (s) φ (s) 1 t φ(s) + 1 t tφ (s) ) dxdt, s = 1,. Direct calculations give us I 1 = t( v v,n ) dxdt and, therefore, I = t v,n dxdt + I. (6.18) Now, our aim is to estimate I from below. Since α ]1/, 1[, we can skip the fist integral in the expression for I. As a result, we have I t v (A 1 + A + A 3 ) dxdt, (6.19) where A 1 = t φ (), For A, we find A 1 t t α xα 4 n a(α 1) A = A 1 φ () 1 t φ(), A 3 = t φ () + 1 t tφ (). [ 4α ax α+ ] n α(α )(α 3). t α+1 Since x n 1 and 0 < t < 1, we see that A > 0 for all a. Hence, it follows from (6.18) and (6.19) that I t v (A 1 + A 3 ) dxdt. (6.0) 40

It is not difficult to check the following inequality On the other hand, A 3 a(α 1) xα n. (6.1) tα+ and thus t φ () 1 t φ() (α 1) 1 t t α+1 4α a x (α 1) n 0 A 1 1 t φ(). (6.) Combining (6.0) (6.), we deduce from (6.5) the estimate t Lv dxdt I a(α 1) x α n v dxdt + t v φ () dxdt t α (6.3) a(α 1) v dxdt + t v φ () dxdt. Using (6.8), we can find the following analog of (6.9) t v dxdt = 1 v dxdt tv Lv dxdt + t v ( φ t φ) dxdt. (6.4) Due to special structure of φ, we have φ t φ = φ (1) t φ (1) + φ () t φ () = φ (1) + φ () t φ () and, therefore, (6.4) can be reduced to the form ( t v + t v ( φ (1) + φ () )) dxdt = ) t ( v + v φ dxdt = 1 v dxdt (6.5) tv Lv dxdt + t v φ () dxdt t v t φ () dxdt. 41

But and, by (6.11) and (6.5), t t φ () a xα n t α 1 te φ u v (tlv) dxdt + t v φ () dxdt + a x α n v dxdt. t α (6.6) The Cauchy-Scwartz inequality, (6.3), and (6.6) imply required inequality (6.1). 7 Appendix Heat Equation We start with derivation of the known estimates for solutions to the Cauchy problem for the heat equation. So, let us consider the following initial problem t u u = 0 in Q T, (7.1) u(, 0) = a( ) in R 3. (7.) Lemma 7.1 For solutions to problem (7.1) and (7.), the following bounds are valid: u(, t) s c 1 (s, s 1 )t 1 l a s1, t > 0, (7.3) for s s 1 u s,l,qt c 1(s, s 1 ) a s1 (7.4) for s > s 1. Here, 1 l = 3 1 ( 1 ). (7.5) s 1 s Remark 7. Estimates (7.4) were proved in [18], Chapter 3, Theorem 9.1. Proof We are not going to prove Lemma 7.1 in full generality. Our aim is just to show how it can be done. First, we note that the solution to the Cauchy problem has the form u(, t) = Γ(, t) a( ), (7.6) 4